calculate-pi-8-pi-6-dx-sin-2x-

Question Number 36427 by prof Abdo imad last updated on 02/Jun/18

c a l c u l a t e \int_{\frac{π}{8}}^{\frac{π}{6}} \frac{d x}{s i n (2 x)}

Commented by abdo mathsup 649 cc last updated on 03/Jun/18

I =_{2 x = u} \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{s i n (u)} \frac{d u}{2}

= \frac{1}{2} \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{d u}{s i n u} a n d c h a n g e m e n t t a n (\frac{u}{2}) = x g i v e

I = \frac{1}{2} \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{1}{\frac{2 x}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}} = \frac{1}{2} \int_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} \frac{d x}{x}

= \frac{1}{2} {[l n ∣ x ∣]}_{\sqrt{2} - 1}^{\frac{1}{\sqrt{3}}} = \frac{1}{2} {- l n (\sqrt{3}) - l n (\sqrt{2} - 1)

= - \frac{1}{2} {\frac{1}{2} l n (3) + l n (\sqrt{2} - 1)} .

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18

\int_{\frac{Π}{8}}^{\frac{Π}{6}} \frac{1 + {t a n}^{2} x}{2 t a n x} d x

\frac{1}{2} ∣ l n (t a n x) ∣_{\frac{Π}{8}}^{\frac{Π}{6}}

= \frac{1}{2} {l n t a n \frac{Π}{6} - l n t a n \frac{Π}{8})